Problem: The lifespans of turtles in a particular zoo are normally distributed. The average turtle lives $93$ years; the standard deviation is $5.6$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a turtle living longer than $109.8$ years.
Explanation: $93$ $87.4$ $98.6$ $81.8$ $104.2$ $76.2$ $109.8$ $99.7\%$ $0.15\%$ $0.15\%$ We know the lifespans are normally distributed with an average lifespan of $93$ years. We know the standard deviation is $5.6$ years, so one standard deviation below the mean is $87.4$ years and one standard deviation above the mean is $98.6$ years. Two standard deviations below the mean is $81.8$ years and two standard deviations above the mean is $104.2$ years. Three standard deviations below the mean is $76.2$ years and three standard deviations above the mean is $109.8$ years. We are interested in the probability of a turtle living longer than $109.8$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the turtles will have lifespans within 3 standard deviations of the average lifespan. The remaining $0.3\%$ of the turtles will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({0.15\%})$ will live less than $76.2$ years and the other half $({0.15\%})$ will live longer than $109.8$ years. The probability of a particular turtle living longer than $109.8$ years is ${0.15\%}$.